Question

If x is real, then $\dfrac{x}{{{x}^{2}}-5x+9}$ must lie between:
a) $\dfrac{1}{11}$ and 1.
b) -1 and $\dfrac{1}{11}$
c) -11 and 1
d) 1 and 11
e) $-\dfrac{1}{11}$ and 1

Answer 1

Hint: Start by letting the expression given in the question to be y. Simplify the equation $y=\dfrac{x}{{{x}^{2}}-5x+9}$ , to get a quadratic equation in terms of x. Finally, take the discriminant of the quadratic equation to be greater than or equal to zero.

Complete step-by-step answer:

To start the question, we let the expression $\dfrac{x}{{{x}^{2}}-5x+9}$ to be equal to y.

$\therefore y=\dfrac{x}{{{x}^{2}}-5x+9}$

$\Rightarrow y\left( {{x}^{2}}-5x+9 \right)=x$

$\Rightarrow y{{x}^{2}}-5xy+9y=x$

$\Rightarrow y{{x}^{2}}-5xy-x+9y=0$

$\Rightarrow y{{x}^{2}}-x\left( 5y+1 \right)+9y=0$

Now we consider y to be constant in the above equation. So, the above equation is a quadratic equation in terms of x. Also, we know that the equation must be satisfied for at least one real value of x as we took the equation to be true by letting y to be equal to the expression $\dfrac{x}{{{x}^{2}}-5x+9}$ .

Therefore, we can deduce that the discriminant of the above quadratic equation must be greater than or equal to zero.

$\therefore {{b}^{2}}-4ac\ge 0$

$\Rightarrow {{\left( 5y+1 \right)}^{2}}-4y\times 9y\ge 0$

Using the formula ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$ , we get

$\Rightarrow 25{{y}^{2}}+1+10y-36{{y}^{2}}\ge 0$

$\Rightarrow -9{{y}^{2}}+10y+1\ge 0$

Multiplying both sides of the inequality by -1.

$9{{y}^{2}}-10y-1\le 0$

The roots of the quadratic equation $-9{{y}^{2}}+10y+1=0$ are given by:

$y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{10\pm \sqrt{100+36}}{2\times (9)}=\dfrac{10\pm \sqrt{136}}{18}$

Therefore, our inequality becomes:

$\Rightarrow \left( y-\left( \dfrac{10-\sqrt{136}}{18} \right) \right)\left( y-\left( \dfrac{10+\sqrt{136}}{18} \right) \right)\ge 0$

$\therefore y\in \left[ \dfrac{10-\sqrt{136}}{18},\dfrac{10+\sqrt{136}}{18} \right]$

We can say that the approximate value of $\sqrt{136}$ is 11.5. So, we get

$\therefore y\in \left[ \dfrac{-1.5}{18},\dfrac{21.5}{18} \right]\in \left( -\dfrac{1}{11},1 \right)$

Therefore, we can say that the answer to the above question is option (e).

Note: Remember, whenever you multiply both sides of the inequality by a negative number, the sign gets reversed. Also, before using the formula of the quadratic equation, we must ensure that the term's coefficient with the highest power of x is positive; this helps prevent mistakes while solving.